Before checking the answer, I answered the exact opposite. My argument is that, when you combine two vectors in order to form another with a linear combination, those two vectors had to be collinear in order for them to produce another vector in the same linear path (=same slope).

Before checking the answer, I answered the exact opposite. My argument is that, when you combine two vectors in order to form another with a linear combination, those two vectors had to be collinear in order for them to produce another vector in the same linear path (=same slope).

How am I wrong?

The general concept is: If you have two non-collinear vectors and all other vectors in the same plane can be expressed as:

where the coefficients a and b are of degree 1 that means linear.

In other words: The expression linear is refering to the degree of the coefficients.

Before checking the answer, I answered the exact opposite. My argument is that, when you combine two vectors in order to form another with a linear combination, those two vectors had to be collinear in order for them to produce another vector in the same linear path (=same slope).

How am I wrong?

Yes, "in order for them to produce another vector in the same linear path" but the question was for them to produce any given vector. For example, the vectors (1, 1) and (2, 2) are collinear and any linear combination of them: a(1,1)+ b(2,2)= (a+ 2b, a+ 2b) is of the form (x,x). The vector (1, 3) cannot be written as a linear combination of them.

But I would have another objection to that answer. Two independent vectors will form a basis in 2 dimensions, but that is not said by the problem. In fact the problem say "two vectors of a basis" which seems to imply that there may be more vectors in that same basis. But certainly any two vectors in a basis must be linearly independent. In order that any vector could be written as a linear combination of them, the vectors must span the space- they must be the only vectors in that basis.